Gibbs free energy and entropy in enzyme catalysis

Gibbs free energy and entropy in enzyme catalysis

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In enzyme catalysis, the formation of the enzyme-substrate complex causes a reduction in the entropy of the system. If entropy reduces, the gibbs free energy will become less negative. How then is enzyme catalysis favorable? Are there any other factors that off-set this?

As @canadianer hinted, the answer is enthalpy.


For binding to be a spontaneous process ΔG must be negative. If the binding of a ligand to a protein decreases system entropy then ΔS is negative and the entropy term -TΔS is positive. However ligand binding involves the formation of various protein-ligand interactions (hydrogen bonds, ionic interactions, van der Waals interactions) all of which make a negative contribution to the change in enthalpy, so ΔH is negative. If |ΔH| > |TΔS| then ΔG will be negative.

This simplified account ignores the parallel changes taking place due to the ligand 'dissociating' from the solvent.

Lecture 35: Enzyme catalysis

Download the video from iTunes U or the Internet Archive.

Topics covered: Enzyme catalysis

Instructor/speaker: aMoungi Bawendi, Keith Nelson

Lecture 1: State of a syste.

Lecture 3: Internal energy.

Lecture 5: Adiabatic changes

Lecture 6: Thermochemistry

Lecture 9: Entropy and the .

Lecture 10: Entropy and irr.

Lecture 11: Fundamental equ.

Lecture 12: Criteria for sp.

Lecture 13: Gibbs free energy

Lecture 14: Multicomponent .

Lecture 15: Chemical equili.

Lecture 16: Temperature, pr.

Lecture 17: Equilibrium: ap.

Lecture 18: Phase equilibri.

Lecture 19: Clausius-Clapey.

Lecture 20: Phase equilibri.

Lecture 21: Ideal solutions

Lecture 22: Non-ideal solut.

Lecture 23: Colligative pro.

Lecture 24: Introduction to.

Lecture 25: Partition funct.

Lecture 26: Partition funct.

Lecture 27: Statistical mec.

Lecture 29: Applications: c.

Lecture 30: Introduction to.

Lecture 31: Complex reactio.

Lecture 32: Steady-state an.

Lecture 33: Chain reactions

Lecture 34: Temperature dep.

Lecture 35: Enzyme catalysis

Lecture 36: Autocatalysis a.

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PROFESSOR: So, let me start here with the temperature dependence of k. And this turns out to be extremely important. And it's due to Arrhenius in 1889. And Mr. Arrhenius is famous for many reasons. Not just for his rate law. So I learned recently, turns out that he was also one of the first people to calculate the potential effect of carbon dioxide in the atmosphere. And and its potential effect on global warming. This was the Industrial Revolution and people were starting to use fossil fuels at increasing amounts. And back then, already, some people started to get worried. And so he did the calculation. It was a very crude calculation. And he extrapolated, assuming that the rate to fossil fuel use would keep going, he got his calculation pretty much right. What he didn't get right was the amount of fossil fuel that people would be using. And so when he did his calculation, which is again a very crude calculation, he got that we would get in trouble at about 2,000 years from the time of his calculation. So he said, no problem, 2,000 years, we've got a lot of time.

And ever since then people have redone these calculations and more and more sophisticated. But his crude calculation was good enough. And as people redo the calculations, the time that they say we're in trouble to the time of the calculation gets closer and closer together. And the reason for that isn't that their calculations are wrong, it's that the rate at which we put out carbon dioxide just keeps getting faster and faster and faster. So it's interesting to go back to these calculations through the last century and a half and see that. Anyway, so Mr. Arrhenius predicted global warming. And he wrote his rate equation, k is equal to e to the minus Ea over RT. Famous Arrhenius rate equation. If you plot it as log of k is equal to log A minus Ea over RT, you find that it looks like a straight line. With an intercept at log A, where A is the pre-factor here and Ea is going to be the activation energy. And the slope is Ea, minus Ea over RT. R is the gas constant. T is the temperature.

And if you put in some typical numbers, activation energies typically are on the order of a few tens to hundreds of kilojoules per mole. Let's say 50 to 300 kilojoules per mole. And typical activation factors, typical pre-exponential factors, are depending if it's a first order or second order. Let's say, first order 10 to the 12th, 10 to the 15th, per second, so the A carries the units for the rate here. So if you have A, so this is for first order, if you have a second order, then the units of A will be different. And typically they're going to be on the order of 10 to the 11th or so, 10 to the 10th, 10 to the 12th, one over molar per second, for second order.

And it's interesting to do sort of a back of the envelope example to figure out how important this rate is. And one of the interesting things to look at is, suppose that I'm at some temperature and I want to know how do I change the temperature. How much do I need to change the temperature to double the rate. And let's say my temperature's at, say, T1 is on the order of 300 degrees Kelvin, room temperature. Room temperature, and let's take a typical Ea on the order of 100 kilojoules per mole. Room temperature, or body temperature are roughly the same, right? These things are.

And so the question is, what does T2 need to be to double the rate connected with an activation energy which is fairly typical of 100 kilojoules per mole. So we want to know, what is, if I have k2 over k1 is equal to two, what's the new T2. Alright so A e to the minus Ea over R T2 divided by A e to the minus Ea over R T1, we want that equal to two, the A's cancel out. So log k2 over k1, which is equal to log two, is equal to Ea over R, one over T1 minus one over T2. We know what this is, we know what this is, we solve for T2 and we find that T2 is 305 degrees kelvin. It's a pretty small change to double the rate.

So it's pretty important that your body temperature doesn't change very much. If you have a fever and your temperature goes up, rates start going up. In your body, you could cause a lot of problems if the rates of some reactions go up by a factor of two. So, very sensitive to temperature. Rates are very sensitive to temperature. What are these things physically? Ea and and the pre-exponent factor. Let's take a look at that.

So physically what's going on is, you have your two, let's say it's a bimolecular reaction, A plus B goes to C. So mechanistically what we think is happening is that the molecules come together and collide, right? You have A and B getting together and colliding. And the hypothesis is that when they collide they form a complex. They sort of bind together momentarily. And form this complex that has all the kinetic energy, or part of the kinetic energy of this collision as part of it. So A and B are bound together in some highly excited complex. Which then falls apart to give the product by rearranging its atoms around.

And if you plot the energy of this process as a function of the reaction, we call this the reaction coordinate, then we're going to start at some delta G, or some energy. For the reactants. We're going to end up with some energy for the products. And it could be higher or lower. Let me make them higher here, just because I don't have enough room on the board. Products. So if this is an endothermic reaction, that's the energy for C, this is the energy for A plus B, and then along the way we have this complex that captures some of the energy of the collision. Up here somewhere. And this is the energy, then, at A B star, which we call the activated complex.

So for the reaction to happen, then, we have to go over the barrier and then back to the product. And this distance from the energy of the reactants to the top of the barrier, that's Ea. That's the energy of activation. How much extra energy you have to put in there to go over the barrier. So it's very clear that as you increase the temperature and you increase the amount of energy that the reactants thermally have, or in terms of their kinetic energy, as you raise the temperature you get more and more kinetic energy, you're going up higher and higher in the Boltzmann distribution. And the number of reactants that can make it over the barrier clearly goes up exponentially. As the temperature goes up.

So this is Ea for the forward reaction. Clearly there's going to be an equivalent activation energy for the backward reaction. The two are related. The difference between the forward and backward reactions, so E backwards minus Ea forward gives you the delta G for the reaction itself. So, Ea backwards minus Ea forwards gives you delta E for the reaction. Where E could be enthalpy or it could be free energy.

So that's the physical origin of this Ea, this activation energy. Now, A, the pre-exponential factor, that's the rate of attempt for the molecules to try to go over the barrier. So it has units that make sense for that. And this A is different than this molecule here. Rate of attempt. How many times per unit time, or how many times per second, do these two molecules try to collide? They never collide, they never try, they'll never make it over. So it's one over second, or one over mole second. So this is the rate of attempt. Then e to the minus Ea over RT is the probability of success, of making it over the barrier. So the rate of attempt times the probability of success gives you the rate per unit time of making it over the barrier. Any questions on the dissection of this Arrhenius rate law?

Alright, so here's some examples, a few examples that you know of. Let's look at OH minus plus methyl bromide. Displacement reaction going to this OH minus attacks the carbon right here. To form an activated complex H, H, H, with the OH coming in like this. Let me go like this. OK, there's the intermediate right here. The activated complex in your reaction. Which then falls apart. It can fall apart two ways. The OH can be spit out again, forming back to the reactants, or the bromine can be spit out, forming the product. Which in this case here is methanol. H, H, H, OH, plus Br minus. Typical example.

So, given this picture here, which you're probably already somewhat familiar with, we can then move on to talk about how to affect this barrier. How to change the rate. Oh, I should say a couple more things. One more thing, too. Now, the rate of, so clearly the Ea's are related. Because the difference of the two Ea's is the delta E for the reaction. But the pre-exponential factors, these A's, for the forward reaction and the backward reactions aren't necessarily related. You can imagine that in one case you have, first of all they don't even have to have the same units. You could have a bimolecular reaction one way. And a unimolecular reaction the other way. So you can't really say anything about the forward versus the backward rate this way. All you know is about the energies.

OK, catalysis. So now we have this barrier we have to overcome. And, so suppose that you have two motivators. Suppose you have an equilibrium case which is slow. And a reaction which is slow in both directions, k1, k minus one. And you don't want to wait for this to happen. And this could be in a biological environment. Or it could be an industrial process, like the Haber process. You don't want to wait. And so one of the ways that you can speed it up, you know is by changing the temperature. You change the temperature, the rate goes up. You change it by five degrees, the rate goes up by a factor of two. You can make a huge change by changing the temperature. But, so if you raise T, speeds up. But, K equilibrium also changes. And we saw that when we did the Haber process. We raised the temperature. The rate speeds up, but the equilibrium switches to the reactants. That's no good. So changing the temperature is not always the best thing to do if you want to change the rate. Instead, what you can do is use a catalyst. if you find one. A molecule that reacts with your reactant to form a product, that's B, spitting back that molecule C again without using it up.

And now with the activation energy, we can understand what the catalyst does. What the catalyst does is speeds up those rates, k2 and k minus one, by changing the activation energy. By making the E smaller. So now I have a catalyst, and I can make this energy smaller. So this would be A plus A C activated, getting ready to spit out B.

So in my example here where I have A plus B goes to products, so I would have A, some combination of A, B, and C together, to give out the products. In this case, the difference in energies doesn't change. All that you're changing is that the hump, in both ways. Equilibrium constant doesn't change. Just the rate changes through the Arrhenius rate law. And this is extremely powerful. Especially in biology. So let me give you some examples here.

This is sort of a typical example of increasing rates. Let's say you have the reaction, hydrogen peroxide, goes to water plus oxygen. If you take a little bit of hydrogen peroxide and you put it on your skin, it starts to bubble. But if you let it sit on the bottle nothing happens to it. Put it on your hair, your hair turns white. Blond. But again, if you just let it sit in the bottle very little happens. Well, it happens but very, very slowly over time.

So if you look at the rate of this reaction here, if the rate, moles per second, with no catalyst at all, the rate is 10 to the minus 8 molar per second. Which is very slow. And the activation energy in kilojoules per mole, in this case here is 71 kilojoules per mole.

Now we can start adding catalysts. We can start adding inorganic catalysts like hydrogen bromide. Increase the rate at 10 to the minus 4. So this creates a complex with the hydrogen peroxide, which lowers the barrier to 50 kilojoules per mole, a small amount of lowering. But because the energy is in the exponent there, it makes a big change in the rate. Yes.

PROFESSOR: The rate is independent of the catalyst concentration.

PROFESSOR: The rate would have to change. You're right. That's a good question and I'm not prepared to answer it. I'm going to have to think about this.

Let's pretend now that we are at per unit concentration of the catalyst. And then, and I'll look into it. OK, so this is what happens with an inorganic catalyst. And instead if you use a generic biological catalyst, an enzyme catalase, it's a sort ubiquitous enzyme which is why you have it on your skin, et cetera, then this rate become 10 to the 7th. And the activation energy drops to eight kilojoules per mole.

So your question really has to do with the units of A. Of the pre-factor right there.

OK, and so there are all these examples of reactions. That are very important biologically. Where with an inorganic catalyst, an organic catalyst, you can change the rate by maybe a few orders of magnitude. But as soon as you put in an enzyme, the rate changes by ten orders of magnitude. Or in this case eight plus seven, fifteen orders of magnitude. Humongous change in the rate.

So you've probably done some enzyme catalysis before. But it's probably a good idea to quickly do it again. Because it's just so important. And it ties together our approximations that we've learned about. So enzyme catalysis, so enzymes can be either, could be heterogeneous catalysis, can be homogeneous catalysis. Enzymes are ubiquitous in the biological environment. They serve to regulate the cellular activity in very complicated ways. The cell will up-regulate or down-regulate the concentration of enzymes as it needs to make more or less products. And there's whole cascades of events that happen in this way. And they're in very small concentrations. But they play an extremely important role. Because they're also extremely specific. So you can have an enzyme that will only act on one part of the biological cycle. And affect it by 10 orders of magnitude or 15 orders of magnitude. But not affect any other protein that's around. And that's amazing. And it turns out that a lot of the diseases of old age like, what I'm about to face, or am facing already, you, not yet, but, are the result of some of these enzyme up-regulation, down-regulation, beginning to break down.

So we have all these reactions going on that need to be essentially perfect. When you have DNA replication or protein folding, or things like this. And errors are made. Errors are made all the time in these processes. And errors cause diseases. And so, even as a baby, your biological processes make errors. But you have these processes, these enzymes especially based on enzymes, that can go in there and sort of fix things. Fix things and make sure that you don't end up getting Alzheimer's at age six months. But as we get older, for some reason, these repair processes lose their bearings. Just like we lose our bearings. And and they can't repair things any more. They don't do it very well. And that causes all sorts of diseases. Cancer is probably one of the diseases, due to the lack of being able to repair problems. Alzheimer's. All sorts of dementias. MS. Just name a chronic disease and it's probably due to a problem with up-regulation or down-regulation of some proteins, some enzymes, that are due to a repair process.

So anyway, these enzymes are big proteins. 10 to the 4th, 10 to the 6th molecular weight proteins. On that order or so. They tend to be fairly large in size. On the order of nanometers. Let's say ten nanometers to 100 nanometers. That could be, that's a little bit on the big side. Probably closer to ten nanometers. Ten nanometers is kind of big, for any sort of biological molecule. And they always end with their name ase. Like catalase, uriase, rnase. Esterase, clips ester bonds. Your liver is full of esterases. Because it likes to break things down into smaller and smaller pieces and lots of ester bonds and things that are not very biologically interesting. And that's one way of the liver breaking things down.

So the way it works is that you have your reactants, which in the biological language are called your substrate, come in, into an enzyme. Gets bound up in a pocket of, there's the substrate, reactant. There's the enzyme here. Forms a complex. And then product gets spit out. And the product floats off. And does its thing. It probably gets bound to another enzyme, which makes another product. Et cetera, and the cascade goes on. The signaling cascade goes on. And this enzyme is in very small concentration. The product goes away, so that's going to stay as a very small concentration as well.

So let's observe experimentally, then. Experimentally, what's seen is that the substrate makes products in the presence of the enzyme. With a rate dP/dt, that's, at t equals zero this is the initial rate is proportional to the concentration of the enzyme. This is initial rate. And in the language of enzymatic kinetics, this is called the velocity. dP/dt is also called the velocity. Velocity, moles per unit time. And this would be called, then, the initial velocity of v initial. v initial's proportional to the enzyme concentration. And what else is seen? For fixed, if I fix my concentration of enzyme, I look at the velocity over time, let's say minus dS/dt, which is dP/dt, I find that that's proportional to S. For small S. Small concentration of substrate. And then at large concentration, it's a constant.

So it's not a straight line. In fact, I don't want to do it on this board here. I'll do it on this board here. Plot the concentration of substrate on this axis, and I plot the velocity, or the rate, of the reaction on that axis here. What I find is that it's a constant. So it's going to saturate, the rate is going to saturate to some number. And it's going to start linear with substrate at the beginning. So it's going to be a straight line at the beginning. And eventually it will saturate. That sort of slope. And the saturation point, that's the maximum velocity, or maximum rate that it can have. So we call this v max. And this part here, the velocity is proportional to S.

And somehow we have to find, explain this. Using what we know from kinetics. So we have to come up with a mechanism, and then solve the mechanism. And make sure that it reproduces the data. So we're not the first ones to do this, obviously. Michaelis and Menton did this many years ago. For this mechanism. And the idea is, you have the enzyme plus a substrate react, k1, k minus one, to form a complex. Enzyme substrate complex. But unlike what we've drawn before in terms of this hump, where there's an activated complex which is not stable, in this case here this enzyme substrate combination lasts a long enough time that it's basically a stable complex.

So we're going to write this as a real intermediate that lasts long enough for you to be able to fish it out. And characterize it. And then eventually, that, k2, k minus two, goes to product plus enzyme. So the enzyme is a catalyst that forms a long-lived complex. And if you were to draw this, then, in our energy diagram, where we have the reaction coordinate here, you start out with your enzyme plus substrate. Go up, and you form your intermediate up here. ES. And I put a little dimple in there, because it's stable enough that it's not on the top of the hump, but it lives long enough. And it comes back down to form the product. Plus the enzyme. Without the enzyme in there, this hump would be way, way up there. Would be maybe a factor of ten higher. So this really lowers it a lot.

Now we can start to solve this mechanism. And I forgot one arrow. There which is the arrow going back. Which you usually don't see, but we might as well keep it there. For the sake of completeness. So what do we know? We know that this intermediate concentration is very small. The enzyme concentration itself is very small. Intermediate is very small. And it doesn't change very much. Not changing much. So that means that we need to use a steady state approximation.

So let's write down the rate for this intermediate, d[ES]/dt. It gets formed through the forward process. I'm going to put my brackets back in because E and ES would look the same otherwise. Gets destroyed. Through the backward ways. And I'm going to use a steady state approximation. So I'm already going to start adding steady state here every time I see an intermediate. Get it destroyed to make products. It gets recreated through the backwards reaction from the products. And I'm going to set that equal to zero for the steady state approximation.

Now, we don't really want to have this E floating around here. Because this is something that is very hard to measure. It's much easier to measure the initial concentration of the enzyme. What you put in there, instead of the amount that's not being bound up. So we're going to solve, instead, in terms of [E] is equal to [E]0 minus [ES], where this is the initial concentration and this is the amount that's binding substrate. And this is the amount of free enzyme here then. And when you do that, and you plug in here, and you plug in here, you get your result. Which is that [E] steady state, [ES] steady state, is this ratio. k1 times [S] plus k minus two times the product divided by k minus one plus k2 plus k1 times [S] plus k minus two times the product, times proportional to the initial concentration of enzyme.

And then you can take your steady state approximation for the intermediate and plug it back into your rate equation. So the velocity, we defined as the rate which is minus d[S]/dt, which is k1 times [E] times [S], this is the destruction of the substrate minus k minus one, times [ES] steady state, the backwards process. So we put in, instead of this [E] we put in [E] is equal to [E]0 minus [ES] steady state. And then we put in for [ES] steady state, we put in what we found here. We turn the crank on the algebra. And we find that the velocity, then, is k1 k2 times the substrate concentration. Minus k minus one times k minus two times the product concentration. The whole thing times the initial enzyme concentration. And then k minus one plus k2 on the bottom. Plus k1 [S] plus k minus two times the product.

So, as I mentioned, this product here usually just floats away. And so locally, where you're doing the, and then it gets used by the next step in the cycle. So this is pretty much zero. We can pretty much ignore this. We can ignore this guy here too. Because it's not, you don't have any steady state or an equilibrium. You have a steady state but not an equilibrium situation. And so in that case here, you can rewrite this then as k1 k2 times the substrate times [E]0 divided by k minus one plus k2 plus k1 times the substrate.

And now we can look at these experimental observations and see whether they match our mechanism. If we can understand something. So let's look at the initial rate. Initial rate is supposed to be proportional to the substrate. Initial rate is supposed to be proportional to the substrate. So the initial rate, k1 k2 plus k1, if at early times, plus one. So the initial rate is going to be, somehow I've got the product missing here. No, I got it backwards here. This is not the initial rate. This is the rate where [S] is small.

The initial rate is when you don't have any products made. Or when the product concentration is very low. And because we're making the assumption that the product concentration is basically equal to zero here, this is basically the initial rate here. So by taking the product equal to zero here, we're also saying that this is the same thing as the initial rate. So this is what it looks like here.. The initial rate here. And this is, this, you rewrite as v initial is equal to k2 times [S] times [E]0. You divide by k1 up and down. And you have this k minus one plus k2 over k1 and then plus [S] sitting down there. And you define this ratio of rates, k minus one plus k2 over k1 as the KM, the Michaelis constant, by definition.

And this is an interesting ratio. This is the rate, k minus one is the rate of destroying the enzyme substrate complex by going back to the reactants. k2 is the destruction of the complex by going to the product. And k1's the creation of the complex. So this is the rate of destruction of the complex divided by the rate of creation of the complex. So if the rate of destruction of the complex is much faster than the rate of creation, meaning that this is a large number, then you're not going to pile up any complex. It's going to be destroyed as soon as you create it. So if KM is large, then the concentration, [ES], is going to be very small. Compared to [E]0. But if the rate of destruction of the complex is small compared to the rate of creation, you create complexes, you create complexes but you don't keep up in terms of destroying them, in terms of making products, or going back to reactants. And so you end up saturating your enzyme. Every enzyme ends up having substrate bound to it. So when KM is very small, then [ES] goes to saturation. Basically, when KM is very small, then you're limited by the rate, the second rate in the process, of the enzyme falling apart. To form the product. You have to wait until that happens. Because then the product just floats away. And that becomes your rate limiting step.

Another way that you also will see this written is as this, then, is equal to, we'll define k cat is equal to k2. k cat times the enzyme concentration times the substrate, [E]0, divided by KM plus but the substrate concentration.

So let's look at a few limiting cases then. First limiting case is, suppose, that [S] is large. Let's take [S] to be much larger than KM. Because you look at the denominator, and you see it's this ratio of rates but there's this concentration that's important here. So in one case KM is going to dominate, and in the other limiting case the substrate concentration is going to dominate. So let's say that the substrate concentration dominates. Meaning K sub M is small. In which case we already saw. If K sub M is small then you reach saturation. And in the equation, then, the velocity, KM is small, [S] is large, the [S]'s cancel out and the velocity is equal to k cat times the initial substrate concentration. Both of these are constants. The velocity is constant. And the rate is constant, it's that limit up here. That's experimentally seen.

And the rate is depending on the initial substrate concentration. Initial enzyme concentration. All the enzymes have a substrate in there. The more enzymes you have to begin with, the more intermediates you're going to have, the faster you going to make products. And then it's going to depend on this on the rate, k cat, which is just k2, which is the rate of formation of products. That becomes the rate limiting step here.

The other special case is if substrate concentration is very small compared to KM. And in that case here, and the other thing that we're going to do is, we're going to, because we now understand this as this maximum rate up there, we're going to call this v max. k cat times [E]0, we're going to call it v max. And so when [S] is very small, v, then, is equal to v max, which is k cat times [E]0. This is v max here now. So we can rewrite this as v max times substrate divided by KM plus the substrate concentration. Another way of writing it. Capital K.

v max times the substrate concentration. And we have KM plus [S]. But [S] is very small. So we drop it. And this is then proportional to the substrate concentration. And then that's small substrate concentration compared to KM, we're sitting right here. Where it's linear, where the rate is linear. And there's a third place on the graph which is interesting. Which is when the substrate concentration is equal to KM. In that case there, you plug [S] equal to KM, and you end up with the velocity then is equal to v max divided by two.

So when [S] is equal to KM, you are halfway up. There's v max over two and there's KM sitting here. When [S] is equal to KM, you're at v max over two. And so enzymes, then, are labeled by their KM's. Because then it becomes very important to know how strongly they bind the substrate. Sometimes you want the enzyme to bind it very strongly. Sometimes you don't, you want it to be fleeting. Depends on the role that the enzyme plays.

Now there's a way to plot this that extracts out these important numbers. k cat and KM. And that's the Lineweaver-Burk plot.. Lineweaver-Burk plot.. And I just looked up this morning to see if Mr. Lineweaver was still alive. And as far as I can tell he's still alive. He was 97, in 2003. So as of 2007 he was still alive. He's getting up there. One of the most cited papers that you have in your notes is the in Jack's, was the paper that showed how to go from this curved line to a straight line by plotting one over v versus [S], instead of v versus [S]. So if you take your equation and massage it, one over v KM over v max times [S], plus one over v max, we haven't done anything except rewrite the equation in terms of one of v versus one over [S]. So it becomes linear. In one over [S], and there's one over v sitting here. And you get a straight line. With an intercept here that's one over v max. And if you keep going, extrapolate out, you get this point here to be minus 1 over KM and the slope is KM over v max. And v max was equal to k cat times [E]0, so you get k cat out of this. So it turned out to be a very useful plot. It's very easy to plot a straight line, especially before computers. In the age of computers. And the referees, there were six referees that got this paper and pretty much turned it down because they didn't think there was any new chemistry in it. Which is true, there's no new chemistry. It's just a way of rewriting the plot. But it was very important nevertheless.

OK, any questions on catalysis? Enzymes? Arrhenius? Alright, next time we'll oscillating reactions and recap the course.

Gibbs Free Energy and Its Formula | Thermodynamic System | Energy Management

After reading this article you will learn about the gibbs free energy and its formula,which is a process-initiating work obtainable from an isothermal,isobaric thermodynamic system.

Gibbs free energy is the measures of “useful” or process-initiating work obtainable from an isothermal, isobaric thermodynamic system. It is the maximum amount of non-expansion work that can be extracted from a closed system this maximum can be attained only in a completely reversible process. When a system changes from a well-defined initial state to a well-defined final state, the Gibbs free energy ∆G equals the work exchanged by the system with its surroundings. The free energy change (∆G) of a reaction determines its spontaneity. A reaction is spontaneous if ∆G is negative (if the free energy of the products is less than the free energy of the reactants).

Where ∆G= change in free energy, ∆G 0 = standard free energy change (with 1 M reac­tants and products, at pH 7), R = gas constant, T = absolute temperature

At equilibrium, ∆G equals zero. Solving for ∆G 0 ‘ yields the relationship at left.

K’eq, the ratio [C][D]/[A][B] at equilibrium, is called the equilibrium constant.

The standard free energy change (∆G 0 ‘) of a reaction may be positive and the actual free energy change (∆G) negative, depending on cellular concentrations of reactants and products. Many reactions for which ∆G 0 ‘ is positive are spontaneous because other reactions cause depletion of products or maintenance of high substrate concentrations.

An equilibrium constant greater than one, (more products than reactants at equilibrium) indicates a spontaneous reaction (negative ∆G°’). Free energy changes of coupled reactions are additive.

Examples of different types of coupling:

Some enzyme-catalyzed reactions can be expressed in two coupled half-reactions, one spontaneous and the other non-spontaneous. The free energy changes of the half-reactions may be added, to yield the free energy of the coupled reaction.

For example, in the reaction catalyzed by the Glycolysis enzyme Hexokinase, the two half-reactions are:

Pi + glucose <—> glucose-6-P+ H2O

ATP + glucose <—>ADP + glucose-6-P

Two separate enzyme-catalyzed reactions occurring in the same cellular compart­ment, one spontaneous and the other non-spontaneous may be coupled by a common inter­mediate (reactant or product). For example, reactions involving pyrophosphates,

Overall reaction A + ATP+ H2O <—> B + AMP + 2Pi ∆G 0 ‘ = (+15-33) = -18 kJ/mol

Pyrophosphate (PPi) is often the product of a reaction that needs a driving force. Its spontaneous hydrolysis, catalyzed by pyrophosphatase enzyme, drives the reaction for which PPi is a product.

Active transport of ions through membrane is coupled to a chemical reaction, e.g., hydrolysis or synthesis of ATP, while the transports of an ion say A + creates a potential dif­ference across the membrane.

The free energy change (electrochemical potential difference) associated with transport of an ion A + across a membrane from side 1 to side 2 is represented as:

Where R = gas constant, T = temperature, Z = charge on the ion, F = Faraday constant, and ∆Ψ = voltage across the membrane.

Each of our cells has an electric potential associated with it. This potential, or voltage, helps to control the migration of ions across the cell membranes. A major example of electri­cal work is in the operation of the nerves. The nerves when get stimulated, they generate an electrical impulse called an action potential which can communicate information to the brain, or carry a signal from brain to a muscle to initiate its movement.

Since free energy changes are additive, the spontaneous direction for the coupled reac­tion will depend on the relative magnitudes of ∆G for the ion flux (∆G varies with the ion gradient and voltage) and ∆G for the chemical reaction (∆G 0 ‘ is negative in the direction of ATP hydrolysis. The magnitude of ∆G also depends on the concentrations of ATP, ADP, and Pi).

Two examples of such coupling are:

The membrane transport requires energy or ATP. The spontaneous ATP hydrolysis (negative ∆G) is coupled to (drives) ion flux against a gradient (positive ∆G).

It takes place in mitochondria, which is also known as the power house of the cell and where the synthesis of ATP occurs. The spontaneous H + flux across a membrane (negative ∆G) is coupled to (drives) ATP synthesis (positive ∆G).

Explain why the entropy increased when the enzyme binds to the substrate? I thought binding to the substrate would reduce the entropy because it is more ordered?

The entropy you refer to is the entropy of activation, #DeltaS^(‡)# , which is a function of the temperature and volume of activation. So if the volume of activation has increased (due to substrate binding to enzyme), the entropy of activation has increased.

Or, upon binding to the enzyme, the enzyme disperses some energy outwards to the substrate (think delocalization and resonance), lowering the energy of the transition state (thereby decreasing the activation energy). This increased energy dispersal indicates greater entropy of activation.

Transition states can be modeled by the Eyring-Polanyi equation:

  • #k# is the rate constant for a process.
  • #kappa# is the transmission factor (i.e. the ratio of transmitted to reflected particles). Usually this is assumed to be #1# , saying that the transition state completely proceeds towards product.
  • #k_B# is the Boltzmann constant.
  • #T# is the temperature.
  • #h# is Planck's constant.
  • #DeltaG^(‡)# is the change in Gibbs' free energy of activation, #G^(‡) - G_R# , where #G_R# is the Gibbs' free energy of the reactants.
  • #R# is the universal gas constant.

Using the isothermal Gibbs equation #DeltaG^(‡) = DeltaH^(‡) - TDeltaS^(‡)# , and assuming #kappa = 1# :

Upon rearranging, we get a linear form of the equation:

Now, let's see what could happen to increase the rate constant, i.e. speed up the reaction.

  • After the substrate binds onto the enzyme, the mechanism becomes more dissociative going away from the transition state, i.e. the volume of the transition state is larger than the volume of the reactants ( #DeltaV^(‡) > 0# ), and the substrate in the activated complex is going to break away from it soon.

This increased dispersion of energy from within the enzyme outwards to the substrate makes #DeltaS^(‡) > 0# . The reaction proceeds on the enzyme more quickly now that the bonds in the reactant are more willing to break apart. That increases the rate.

When the enzyme binds (or anything, really), energy is released, so the enthalpy of activation is negative ( #DeltaH^(‡) < 0# ), thereby increasing the right-hand side of the equation, so #k# increases.

Energy and enzymes

One way we can see the Second Law at work is in our daily diet. We eat food each day, without gaining that same amount of body weight! The food we eat is largely expended as carbon dioxide and heat energy, plus some work done in repairing and rebuilding bodily cells and tissues, physical movement, and neuronal activity.

Although living organisms appear to reduce entropy, by assembling small molecules into polymers and higher order structures, this work releases waste heat that increases the entropy of the environment.

Gibbs Free Energy

Gibbs free energy is a measure of the amount of work that is potentially obtainable. Instead of absolute quantities, what is usually measured is the change in free energy:

where H = enthalpy (the heat energy content) T = absolute temperature (Kelvin), and S = entropy (sometimes called disorder, but a complicated and subtle concept that has more to do with degrees of freedom 6 molecules of CO2 have greater entropy than a molecule of glucose, where the carbon atoms are linked together by covalent bonds).

If ΔG < 0, a chemical reaction is exergonic, releases free energy, and will progress spontaneously, with no input of additional energy (though this does not mean that the reaction will occur quickly – see the discussion about reaction rates below).

If ΔG > 0, a chemical reaction is endergonic, requires or absorbs an input of free energy, and will progress only if free energy is put into the system else the reaction will go backwards.

Free energy and chemical equilibrium

If ΔG = 0, a chemical reaction is in equilibrium, meaning that the rates of forward and reverse reactions are equal, so there is no net change, or no potential for doing work.

This figure from Wikipedia illustrates that reactions will proceed spontaneously towards equilibrium, in either direction, and that the equilibrium point is the minimum free energy state of the reaction mixture. The x-axis (ξ) represents the ratio of products/reactants.

Cells couple exergonic reactions to endergonic reactions so that the net free energy change is negative

ATP is the primary energy currency of the cell cells accomplish endergonic reactions such as active transport, cell movement or protein synthesis by tapping the energy of ATP hydrolysis:

ATP -> ADP + Pi (Pi = PO4, inorganic phosphate) ΔG = -7.3 kcal/mol

Cycle of ATP hydrolysis to ADP and phosphorylation of ADP to ATP. The majority of endergonic reactions in cells are coupled to the exergonic hydrolysis of ATP to ADP. Image by Muessig retrieved from Wikimedia Commons, licensed CC-BY-SA 3.0

Over the next pages we’ll be looking at the cellular metabolic pathways that phosphorylate ADP to make ATP.

Reaction rates
Although the sign of ΔG (negative or positive) determines the direction that the reaction will go spontaneously, the magnitude of ΔG does not predict how fast the reaction will go.
The rate of the reaction is determined by the activation energy (the energy required to attain the transition state) barrier Ea:

Energy diagram of enzyme-catalyzed and uncatalyzed reactions, from Wikipedia

The peak of this energy diagram represents the transition state: an intermediate stage in the reaction from which the reaction can go in either direction. Reactions with a high activation energy will proceed very slowly, because only a few molecules will obtain enough energy to reach the transition state – even if they are highly exergonic. In the figure above, the reaction from X->Y has a much greater activation energy than the reverse reaction Y->X. Starting with equal amounts of X and Y, the reaction will go in reverse.

The addition of a catalyst (definition: an agent that speeds up the rate of a reaction, but is not consumed or altered by the reaction) provides an alternative transition state with lower activation energy. What this means is that the catalyst physically ‘holds’ the substrates in a physical conformation that makes the reaction more likely to proceed. As a result, in the presence of the catalyst, a much higher percentage of molecules (X or Y) can acquire enough energy to attain the transition state, so the reaction can go faster, in either direction. Note that the catalyst does not affect the overall free energy change of the reaction. Starting with equal amounts of X and Y, the reaction diagrammed above will still go in reverse, only faster, in the presence of the catalyst.

Enzymes speed up reactions by lowering the activation energy barrier

Enzymes are biological catalysts, and therefore not consumed or altered by the reactions they catalyze. They repeatedly bind substrate, convert, and release product, for as long as substrate molecules are available and thermodynamic conditions are favorable (ΔG is negative the product/substrate ratio is lower than the equilibrium ratio). Most enzymes are proteins, but several key enzymes are RNA molecules (ribozymes). Enzymes are highly specific for their substrates. Only molecules with a particular shape and chemical groups in the right positions can interact with amino acid side chains at the active site (the substrate-binding site) of the enzyme.

Enzyme-catalyzed reactions have saturation kinetics

The velocity of enzyme-catalyzed reactions increases with the concentration of substrate. However, at high substrate concentrations, the quantity of enzyme molecules becomes limiting as every enzyme molecule is working as fast as it can. At saturation, further increases in substrate concentrations have no effect the only way to increase reaction rates is to increase the amount of enzyme.

Enzyme kinetics plot by Thomas Shafee, CC-BY-SA 4.0 from Wikimedia Commons

The kinetic properties of enzymes are defined by their Vmax and Km

  • The Vmax is the maximum rate at which enzymes can work, at saturating concentrations of substrate.
  • The Km (Michaelis constant), is defined as the substrate concentration that produces 1/2 Vmax, and is a measure of the affinity of the enzyme for its substrate.

Enzyme inhibitors
Enzymes are subject to regulation, and are the targets of many pharmaceutical drugs, such as non-steroidal pain relievers. Many enzymes are regulated by allosteric regulators which bind at a site distinct from the active site.

Noncompetitive inhibitors act allosterically (bind at a site different from the active site). When the noncompetitive inhibitor binds allosterically, it often changes the overall shape of the enzyme, including the active site, so that substrates can no longer bind to active site.

Competitive inhibitors compete with the substrate for binding to the active site the enzyme cannot carry out its normal reaction with the inhibitor, because the inhibitor physically blocks the substrate from binding the active site.
My lecture videos on thermodynamics and enzymes (have audio lag, plan to re-do in shorter segments)


As we can see from Equation 9, we have reached, in a formal way, a conclusion that we knew by common sense. That is, under conditions of initial velocity, where the reaction spontaneously evolves from S to P, the free energy change for the formation of the ES complex is always negative. Another interesting conclusion that can be derived from Equation 9 is that by making k−1k2 (the Michaelis-Menten condition for rapid equilibrium) ΔGgs tends toward zero, thus decreasing the free activation energy for the catalyzed reaction and increasing the reaction rate (Fig. 3).

German-British medical doctor and biochemist Hans Krebs' 1957 book Energy Transformations in Living Matter (written with Hans Kornberg) [1] was the first major publication on the thermodynamics of biochemical reactions. In addition, the appendix contained the first-ever published thermodynamic tables, written by Kenneth Burton, to contain equilibrium constants and Gibbs free energy of formations for chemical species, able to calculate biochemical reactions that had not yet occurred.

Non-equilibrium thermodynamics has been applied for explaining how biological organisms can develop from disorder. Ilya Prigogine developed methods for the thermodynamic treatment of such systems. He called these systems dissipative systems, because they are formed and maintained by the dissipative processes that exchange energy between the system and its environment, and because they disappear if that exchange ceases. It may be said that they live in symbiosis with their environment. Energy transformations in biology are dependent primarily on photosynthesis. The total energy captured by photosynthesis in green plants from the solar radiation is about 2 x 10 23 joules of energy per year. [2] Annual energy captured by photosynthesis in green plants is about 4% of the total sunlight energy that reaches Earth. The energy transformations in biological communities surrounding hydrothermal vents are exceptions they oxidize sulfur, obtaining their energy via chemosynthesis rather than photosynthesis.

The field of biological thermodynamics is focused on principles of chemical thermodynamics in biology and biochemistry. Principles covered include the first law of thermodynamics, the second law of thermodynamics, Gibbs free energy, statistical thermodynamics, reaction kinetics, and on hypotheses of the origin of life. Presently, biological thermodynamics concerns itself with the study of internal biochemical dynamics as: ATP hydrolysis, protein stability, DNA binding, membrane diffusion, enzyme kinetics, [3] and other such essential energy controlled pathways. In terms of thermodynamics, the amount of energy capable of doing work during a chemical reaction is measured quantitatively by the change in the Gibbs free energy. The physical biologist Alfred Lotka attempted to unify the change in the Gibbs free energy with evolutionary theory.

Energy transformation in biological systems Edit

The sun is the primary source of energy for living organisms. Some living organisms like plants need sunlight directly while other organisms like humans can acquire energy from the sun indirectly. [4] There is however evidence that some bacteria can thrive in harsh environments like Antarctica as evidence by the blue-green algae beneath thick layers of ice in the lakes. No matter what the type of living species, all living organisms must capture, transduce, store, and use energy to live.

The relationship between the energy of the incoming sunlight and its wavelength λ or frequency ν is given by

where h is the Planck constant (6.63x10 −34 Js) and c is the speed of light (2.998x10 8 m/s). Plants trap this energy from the sunlight and undergo photosynthesis, effectively converting solar energy into chemical energy. To transfer the energy once again, animals will feed on plants and use the energy of digested plant materials to create biological macromolecules.

Thermodynamic Theory of Evolution Edit

The biological evolution may be explained through a thermodynamic theory. The four laws of thermodynamics are used to frame the biological theory behind evolution. The first law of thermodynamics states that energy can not be created or destroyed. No life can create energy but must obtain it through its environment. The second law of thermodynamics states that energy can be transformed and that occurs everyday in lifeforms. As organisms take energy from their environment they can transform it into useful energy. This is the foundation of tropic dynamics.

The general example is that the open system can be defined as any ecosystem that moves toward maximizing the dispersal of energy. All things strive towards maximum entropy production, which in terms of evolution, occurs in changes in DNA to increase biodiversity. Thus, diversity can be linked to the second law of thermodynamics. Diversity can also be argued to be a diffusion process that diffuses toward a dynamic equilibrium to maximize entropy. Therefore, thermodynamics can explain the direction and rate of evolution along with the direction and rate of succession. [5]

First Law of Thermodynamics Edit

The First Law of Thermodynamics is a statement of the conservation of energy though it can be changed from one form to another, energy can be neither created nor destroyed. [6] From the first law, a principle called Hess's Law arises. Hess’s Law states that the heat absorbed or evolved in a given reaction must always be constant and independent of the manner in which the reaction takes place. Although some intermediate reactions may be endothermic and others may be exothermic, the total heat exchange is equal to the heat exchange had the process occurred directly. This principle is the basis for the calorimeter, a device used to determine the amount of heat in a chemical reaction. Since all incoming energy enters the body as food and is ultimately oxidized, the total heat production may be estimated by measuring the heat produced by the oxidation of food in a calorimeter. This heat is expressed in kilocalories, which are the common unit of food energy found on nutrition labels. [7]

Second Law of Thermodynamics Edit

The Second Law of Thermodynamics is concerned primarily with whether or not a given process is possible. The Second Law states that no natural process can occur unless it is accompanied by an increase in the entropy of the universe. [8] Stated differently, an isolated system will always tend to disorder. Living organisms are often mistakenly believed to defy the Second Law because they are able to increase their level of organization. To correct this misinterpretation, one must refer simply to the definition of systems and boundaries. A living organism is an open system, able to exchange both matter and energy with its environment. For example, a human being takes in food, breaks it down into its components, and then uses those to build up cells, tissues, ligaments, etc. This process increases order in the body, and thus decreases entropy. However, humans also 1) conduct heat to clothing and other objects they are in contact with, 2) generate convection due to differences in body temperature and the environment, 3) radiate heat into space, 4) consume energy-containing substances (i.e., food), and 5) eliminate waste (e.g., carbon dioxide, water, and other components of breath, urine, feces, sweat, etc.). When taking all these processes into account, the total entropy of the greater system (i.e., the human and her/his environment) increases. When the human ceases to live, none of these processes (1-5) take place, and any interruption in the processes (esp. 4 or 5) will quickly lead to morbidity and/or mortality.

Gibbs Free Energy Edit

In biological systems, in general energy and entropy change together. Therefore, it is necessary to be able to define a state function that accounts for these changes simultaneously. This state function is the Gibbs Free Energy, G.

  • H is the enthalpy (SI unit: joule)
  • T is the temperature (SI unit: kelvin)
  • S is the entropy (SI unit: joule per kelvin)

The change in Gibbs Free Energy can be used to determine whether a given chemical reaction can occur spontaneously. If ∆G is negative, the reaction can occur spontaneously. Likewise, if ∆G is positive, the reaction is nonspontaneous. [9] Chemical reactions can be “coupled” together if they share intermediates. In this case, the overall Gibbs Free Energy change is simply the sum of the ∆G values for each reaction. Therefore, an unfavorable reaction (positive ∆G1) can be driven by a second, highly favorable reaction (negative ∆G2 where the magnitude of ∆G2 > magnitude of ∆G1). For example, the reaction of glucose with fructose to form sucrose has a ∆G value of +5.5 kcal/mole. Therefore, this reaction will not occur spontaneously. The breakdown of ATP to form ADP and inorganic phosphate has a ∆G value of -7.3 kcal/mole. These two reactions can be coupled together, so that glucose binds with ATP to form glucose-1-phosphate and ADP. The glucose-1-phosphate is then able to bond with fructose yielding sucrose and inorganic phosphate. The ∆G value of the coupled reaction is -1.8 kcal/mole, indicating that the reaction will occur spontaneously. This principle of coupling reactions to alter the change in Gibbs Free Energy is the basic principle behind all enzymatic action in biological organisms. [10]

For Students & Teachers

For Teachers Only


The highly complex organization of living systems requires constant input of energy and the exchange of macromolecules.


Explain how enzymes affect the rate of biological reactions.


The structure and function of enzymes contribute to the regulation of biological processes —

  1. Enzymes are biological catalysts that facilitate chemical reactions in cells by lowering the activation energy.


Understanding the structure and motion of proteins lies at the heart of our understanding of how enzymes achieve spectacular rate enhancements [1] . In recent years, detailed computational studies have clearly demonstrated the importance of the electrostatic environment [2] . Enzymes appear to preorganise the electrostatics of the active site to promote formation of the transition state [3] . Similarly, innovative experimental approaches have demonstrated the importance of conformational sampling of proteins in mediating substrate binding and formation of the reactive complex geometry that precedes the chemical step [4-6] . However, the presence and effect of conformational sampling on enzyme turnover is highly contentious. The free energy landscape (FEL) for an enzyme is a description of protein structure as a multidimensional energy surface, composed of a series of ‘hills’ and ‘valleys’ separating energetic minima. Discrete conformational substates lie within these minima. Alternative substates are accessed by crossing the energetic barriers between minima or altering the FEL itself. That is, the FEL describes the equilibrium of conformational states a protein can adopt. Vibrational free energy contributes very significantly to the shape of the FEL by affecting the distribution of conformational substates within this equilibrium [7] . The frequency of vibrational modes and their population distribution affects the ruggedness of the FEL [7] . For enzyme turnover, it is the specific features of the FEL that affect the rate of enzyme turnover that are important. However, accessing this level of information in a quantitative manner is typically the preserve of high-level simulation studies.

Defining the thermodynamics of an enzyme with respect to its chemical turnover can be achieved by monitoring the change in observed rate in a matrix of pressure/temperature conditions. Numerically fitting these data then gives ΔH ‡ , ΔS ‡ , ΔG ‡ , , ΔV ‡ , Δβ ‡ and Δα ‡ reflecting the changes in enthalpy, entropy, Gibbs free energy, heat capacity, activation volume, compressibility and expansivity between the enzyme–substrate complex and the enzyme–transition state complex respectively. The enzyme kinetics for only a relatively few enzymes has been treated by combined p/T studies [8-10] . In these cases, the p/T plane has not been fitted with a numerical model meaning at least and ∆α ‡ are not determined. We have recently explored the use of a new model that defines the temperature dependence of enzyme-catalysed rates, and that we have termed as macromolecular rate theory (MMRT), (1) where T0 is an arbitrary reference temperature. determines the change in ΔH ‡ and ΔS ‡ with temperature and defines the temperature dependence of the Gibbs free energy between the ground state and the transition state (ΔG ‡ ). The dominant contribution to for enzymes is the distribution and frequency [11, 12] of the large number of vibrational modes of the molecule and its closely associated solvent molecules. Finding that is non-zero for enzyme catalysis demonstrates (in the absence of other contributing factors [13] ) a change in the distribution of vibrational modes between the enzyme–substrate complex and the enzyme–transition state complex. Equivalently, a negative value of implies that <(δH) 2 > (the mean squared distribution of enthalpies) for the enzyme–substrate complex is greater than <(δH ‡ ) 2 > for the enzyme–transition state complex at a given temperature. The magnitude of can therefore be used as an excellent proxy for changes to the enzyme FEL, and more specifically for the changes in vibrational modes during enzyme turnover (conformational sampling). Pressure effects are system specific [9] but should perturb a large as pressure affects the pre-existing equilibrium of conformational states, favouring smaller volumes [14, 15] .

Hobbs et al. have studied a model sugar cleaving enzyme oligo-1,6-glucosidase 1 (MalL EC3.2.1.10) and have found that the wild-type (WT) enzyme displays a significant negative ( = −11.6 kJ·mol −1 ·K −1 ) that is dramatically reduced in the V200S variant enzyme ( = −5.9 kJ·mol −1 ·K −1 ) as well as a significant increase in the optimum temperature (Topt) where kcat is maximal. These changes are not accompanied by a structurally different enzyme form, with the X-ray crystal structures of the WT and V200S enzymes being essentially invariant [16] . Molecular dynamics simulations suggest that the origin of the difference in is due to rigidification of the ground state relative to the transition state for V200S MalL when compared to the WT enzyme [16] . That is, the FEL for the enzyme–substrate complex is greatly constrained at the transition state implying significant changes to the distribution of vibrational modes along the reaction coordinate. WT MalL and the V200S variants therefore present a powerful model system with which to probe the distribution of vibrational modes during enzyme turnover and the relationship to the protein FEL.

Herein, we compare WT and V200S MalL and find that is only pressure-dependent for the WT enzyme and that this is correlated with a more flexible protein, with a broader equilibrium of protein conformational states in the enzyme–substrate complex a broader FEL. Our findings suggest that the vibrational modes of the protein do significantly affect enzyme turnover and that this can be achieved through modulation of Δα ‡ . V200S has higher Topt and a faster rate at Topt. Therefore, our findings suggest that, at least in this case, the faster enzyme has a restricted FEL that minimises and thus changes the temperature dependence of the rate. Contrary to many assertions, for at least some enzymes, decreasing the contribution from conformational sampling for the enzyme–substrate complex is a potential path to achieving more effective (faster) enzymes.

Gibbs free energy and entropy in enzyme catalysis - Biology

Much of what we will study in this course involves reaction pathways--the conversion of one compound to another and another, onward toward some final product that gets used in a structural or functional way in an organism. In order to understand whether the reactions that produce the intermediate and final products will proceed, we need to know whether the reactions give off free energy or require free energy under physiological conditions. If they give off free energy, they will proceed without prompting if they require energy, we will need to understand where the energy to drive the reaction comes from. This is the stuff of thermodynamics.

We observed on Tuesday that thermodynamics alone will not tell us whether a reaction will proceed in a reasonable time frame. If the activation energy that separates reactants from products is high enough, the time required for a system to come to equilibrium will be long enough that the reaction will not play out within the lifetime of an organism. It is the job of biological catalysts--enzymes--to reduce this activation energy enough to make the kinetics of a reaction practical. Kinetics involves an understanding of energy, just as thermodynamics does. Thus the application of energetic considerations in biochemistry involves more than thermodynamics--it involves kinetics, and the ways that enzymes modify kinetics. But for today we'll focus on the energetics of equilibrium, i.e. thermodynamics.

The laws of thermodynamics

  • The first law of thermodynamics says that the energy of a closed system is constant.
  • The second law of thermodynamics says that the entropy in a closed system increases.


Thermodynamic properties



We've already said that entropy is a measure of the disorder in a system. Entropy turns out to be proportional to the logarithm of the number of degrees of freedom &Omega in a system:
S = k ln &Omega
where k is Boltzmann's constant, 3.4*10 -24 cal/deg K, or 1.38*10 -23 joule/deg K.. We often measure entropy in entropy units eu = 1 cal/deg K. The gas constant R is the product of Avogadro's number N and k, so if the entropy of a single molecule is S, then the entropy of a mole of the same kind of molecules will be NS = R ln Ω.
The second law of thermodynamics says that in general the entropy of a closed system will increase, i.e. that for the most part the universe tends toward a larger number of degrees of freedom or a larger amount of disorder.
The entropy of single molecule can be characterized by statistical-mechanical methods if the molecule is simple enough. The following table, adapted from table 2.1 in Zubay's Principles of Biochemistry, breaks the entropy of liquid propane into translational, rotational, vibrational, and electronic components:

type of entropy
This pattern, in which most of the entropy is translational and rotational, is typical of biomolecules. By contrast, the enthalpy in a biomolecule is usually dominated by electronic properties. Translational entropy depends primarily on (3/2)RlnM r , where M r is the molecular weight. In a dimerization reaction the total entropy decreases, because M r doubles, but the logarithm of it does not double--it only increases by ln 2. Thus the translational entropy goes down

Rigidity decreases entropy, because rigid structures cannot rotate as freely and often cannot vibrate as freely.

Entropy in solvation and binding to surfaces

What happens when molecules go into solution? The solute molecules usually undergo an increase in entropy, because they become free to dissociate from one another, and in the case of ionic solutes the cations can separate from the anions. On the other hand, the solvent molecules frequently become more organized in the vicinity of the solute molecules than they had been before the introduction of the solute, so their contribution to total change in entropy is frequently negative. The net effect is often slightly negative, i.e. the solution has a slightly lower entropy than the separated components.

When an apolar molecule is added to water, the water molecules often form a micelle around the foreign molecule. This micelle is highly ordered, so the entropy of the system decreases.

Many biochemical reactions involve binding of small molecules to a surface, e.g. the surface of a protein. In inorganic chemistry the binding of small molecules to surfaces often involves a decrease in entropy because the molecules binding to the surface lose rotational degrees of freedom. But in biochemistry the loss in rotational freedom is more than compensated for by the increase in entropy associated with the release of water molecules from the protein surface. Thus the binding of metabolites to a protein is often entropically favored.

Free energy

Josiah Gibbs articulated the concept of free energy (sometimes called Gibbs free energy), which is related to entropy and enthalpy by
G = H - TS
The change in free energy when a reaction occurs is
&DeltaG = &DeltaH - T&DeltaS
assuming the temperature does not change. Temperature in a biochemical system in general changes very slowly, so this is a reasonable assumption. Gibbs was able to show that a chemical reaction will occur spontaneously if and only if the change in free energy is negative:
&DeltaG < 0
For the most part we will analyze biochemical reactions in terms of their spontaneity and therefore in terms of whether &DeltaG < 0.

We can compute &DeltaG per mole for a wide variety of compounds. A useful formulation is that of the standard free energy of formation of a compound &DeltaG o f, which according to Zubay is the difference between the free energy of the compound in its standard state and the total free energies of the elements of which the compound is composed. This table (again adapted from Zubay) contains some examples of &DeltaG o f values for metabolites:

&DeltaG o f, kcal/mol
&DeltaG o f, kJ/mol
lactate, 1M
pyruvate, 1M
succinate, 1M
glycerol, 1M
acetate, 1M
oxaloacetate, 1M
hydrogen ions, 10 -7 M
carbon dioxide
bicarbonate, 1M
We can use these values to calculate the overall change in standard free energy &DeltaG o associated with a biochemical reaction. There are some tricks and special cases to consider. But the concept is straightforward: given the known values of &DeltaG o f for the reactants and products in a reaction, we can calculate the overall change in standard free energy in a reaction by adding up the &DeltaG o f values for the products and subtracting the &DeltaG o f values for the reactants. The &DeltaG o f values are generally negative, so we'll be subtracting a negative number from another negative number. If the total comes out negative, the reaction is spontaneous if it comes out positive, the reaction is not spontaneous.

Free energy and equilibrium

Free energy as a source of work

The change in free energy tells us the maximum amount of useful work that can be derived from a biochemical reaction. If &DeltaG o is negative, then the largest amount of useful work that could be extracted from the reaction is -&DeltaG o . Some of that energy will go into heat, though, so the actual amount of work we can get will always be less than -&DeltaG o .

  • To move objects, as in muscle contraction and flagellar swimming.
  • To move molecules against concentration gradients and ions across potential gradients.
  • To drive otherwise endoergic reactions either by direct coupling or by depleting concentrations of reactants enough to make the reaction favorable.

Coupled reactions

In some cases, a single enzyme catalyzes two successive reactions, the first of which is exergonic and the second of which is endergonic. In that case, in effect, the overall reaction happens in one shot, with the energy from the exergonic part of the sequence driving the energonic part. If the overall &DeltaG o < 0 for the pair of reactions, the products will be produced.

In other cases, two reactions may not be spatially coupled. Instead, the fact that the first reaction produces a high concentration of its product(s) results in a high concentration of the reactant(s) for the second reaction. Based on the definition of Keq, this imbalance in concentration changes the value of &DeltaG enough to render the second reaction possible.

ATP as an energy currency

The most common compound involved in this process is adenosine triphosphate. It can be hydrolyzed at either the gamma phosphate (the one farthest from the ribose ring) or at the beta phophate (the one in the middle). In the former case, about 7.8 kcal/mol is released by the hydrolysis:
ATP + H2O -> ADP + Pi
where Pi is a standard abbreviation for inorganic phosphate, i.e. PO4 -3 , HPO4 -2 , or H2PO4 - . A similar amount is released in hydrolysis at the beta phosphate:
ATP + H2O -> AMP + PPi,
where PPi is a standard abbreviation for inorganic pyrophosphate, i.e. P2O7 -4 , HP2O7 -3 , or H2PO7 -2 , or H3PO7 - . However, pyrophosphate hydrolyzes into two molecules of ordinary phosphate, with the release of a similar amount of energy. Appropriately coupled, the hydrolysis of ATP to AMP and two equivalents of Pi can therefore yield nearly 16 kcal/mol of energy--enough to drive almost all conventionally-encountered biochemical reactions.

ATP thus acts as a kind of energy currency: a means of storing energy that can be tapped for driving endergonic reactions to completion. The energy has to come from somewhere: it comes from the creation of ATP, with its high-energy phosphate bonds, from lower-energy substituents, using various exergonic reactions as drivers. We can think of the resting concentration of ATP in a cell as the equivalent of a roll of quarters that the cell can spend when it needs energy. Each ATP molecule acts as a single quarter when it's hydrolyzed to ADP it acts as a pair of quarters when it's hydrolyzed to AMP. Most of the purchases the cell needs to make are for prices either just below .50 (ATP -> AMP) or just below .25 (ATP -> ADP), so it's useful currency for the cell to carry around. None of the cellular vendors gives change, so if we use our quarters to buy .03 worth of merchandise at a time, it's not very cost-effective, but in buying items that cost .24 or .48, they're pretty efficient. When the cell runs out of quarters, it needs to go to the metabolic bank and get some more.

Other high-energy compounds

There are other compounds employed as energy-storage entities in cells. None of the others is as plentiful or as widely-used as ATP, but they play significant roles in certain pathways. Each of these compounds contains at least one high-energy phosphorus-oxygen bond, just as ATP does, so the mechanisms are similar to those found in ATP hydrolysis. But the specific &DeltaG values for each of these phosphate compounds differs from that of ATP, and as such they turn out to be more efficient in driving particular classes of reactions. So the cell may be carrying around several rolls of quarters (ATP molecules), but it also carries around one roll of 40-cent pieces (creatine phosphate), one roll of 35-cent pieces (phosphoenol pyruvate), and the like. Since the vendors don't make change, creatine phosphate is a useful compound to carry when making 38-cent purchases.

Watch the video: Gibbs Free Energy, Entropy and Enthalpy. Biochemistry (August 2022).